The polynomial $p(x)=3x^3-20x^2+37x-20$ has a known factor of $(x-4)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Solution: We know $(x-4)$ is a factor of $p(x)$. This means that $p(x)=(x-4)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x-4)$ $\begin{array}{r} 3x^2-\phantom{1}8x+\phantom{1}5 \\ x-4|\overline{3x^3-20x^2+37x-20} \\ \mathllap{-(}\underline{3x^3-12x^2\phantom{+37x-20}\rlap )} \\ -8x^2+37x-20 \\ \mathllap{-(}\underline{-8x^2+32x\phantom{-20}\rlap )} \\ 5x-20 \\ \mathllap{-(}\underline{5x-20\rlap )} \\ 0 \end{array}$ We find that $q(x)=3x^2-8x+5$. Factoring $q(x)$ We can factor $q(x)$ by grouping: $\begin{aligned} q(x)&=3x^2-8x+5 \\\\ &=3x^2-3x-5x+5 \\\\ &=3x(x-1)-5(x-1) \\\\ &=(3x-5)(x-1) \end{aligned}$ Putting it all together $\begin{aligned} p(x)&=3x^3-20x^2+37x-20 \\\\ &=(x-4)(3x^2-8x+5) \\\\ &=(x-4)(3x-5)(x-1) \end{aligned}$